Use the boxplot to identify the maximum value, minimum value, median, first quartile, third quartile, and interquartile range. a. 500, 400, 450, 350, 475, 125 b. 500, 350, 450, 400, 475, 3 c. 475, 400, 450, 350, 500, 150 d. 500, 350, 450, 400, 475, 75

d. 500, 350, 450, 400, 475, 75

Step-by-step explanation:

Hello!

In the graph, the box is delimited by the 1st and 3rd Quartiles, its width represents the IQD and the 2nd Quartile is always the line drawn inside the box.

Then the side of the box is the 1st Quartile and the right side of the box is the 3rd Quartile, the 2dn Quartile (Median) is inside the box:

1stQ= 400

2ndQ= 450

3rdQ= 475

IQD= 3rgQ-1stQ= 475-400= 75

The whiskers of the box plot are a line drawn from the box to the min/max value.

The left whisker is the line drawn from the 1st Quartile to the minimum value of the sample and the right whisker is the line drawn from the 3rd Quartile to the maximum value of the sample.

With this in mind you can identify the min and max values in the graphic:

Min= 350

Max= 500

I hope this helps!

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3